# D.2 Electric fields

> IB Physics HL · Theme D: Fields
> Source: https://www.owlsprep.com/study/ib-physics-hl-u4-d-2-electric-fields/

This sub-topic covers the definition and properties of static electric fields, including uniform fields between parallel plates and radial fields around point charges. You will learn to calculate field strength and solve force and motion problems.

**Prerequisites:** [Coulomb's law for point charges](https://www.owlsprep.com/study/ib-physics-hl-u4-d-1-describing-fields/); Vector addition and Newton's laws of motion

## Learning objectives

- Define electric field strength in terms of force on a test charge
- Calculate electric field strength for radial and uniform electric fields
- Apply kinematics to find motion of charged particles in uniform fields
- Use superposition for net fields from multiple point charges

## Definition of Electric Field Strength

An electric field is a region of space where a stationary electric charge experiences a force. It describes how the electrostatic force acts on charges at any point in the region, and follows the principle of superposition for multiple sources.

**Electric field strength** — Force per unit positive test charge at a point, defined as $E = \frac{F}{q}$, where $F$ is the force on test charge $q$. Units are newtons per coulomb (N C⁻¹) or equivalent volts per metre (V m⁻¹).

*Notation:* E

**Worked example:** A test charge of $+1.5 \times 10^{-6}$ C experiences an upward electrostatic force of $4.5 \times 10^{-3}$ N in an electric field. Calculate the magnitude and direction of the field strength.

1. Use the definition of electric field strength:
2. $$E = \frac{F}{q} = \frac{4.5 \times 10^{-3}\ \text{N}}{1.5 \times 10^{-6}\ \text{C}} = 3000\ \text{N C}^{-1}$$
3. By definition, electric field direction matches the direction of force on a positive test charge, so the field points upwards.

## Radial Electric Fields from Point Charges

A single point charge $Q$ produces a radial electric field, where field strength decreases with the square of distance from the charge. This formula is derived directly from Coulomb's law: $F = \frac{kQq}{r^2}$, so dividing by $q$ gives the field strength.

> **info**
>
> Direction: For a positive source charge, the field points away from the charge. For a negative source charge, the field points towards the charge.

**Worked example:** Calculate the electric field strength 0.10 m away from a point charge of $-2.0 \times 10^{-6}$ C, given $k = 9.0 \times 10^9$ N m² C⁻².

1. Use the radial field formula:
2. $$E = \frac{kQ}{r^2} = \frac{(9.0 \times 10^9)(-2.0 \times 10^{-6})}{(0.10)^2}$$
3. $$E = -1.8 \times 10^6\ \text{N C}^{-1}$$
4. The negative sign indicates the field points towards the negative charge, with magnitude $1.8 \times 10^6$ N C⁻¹.

## Uniform Electric Fields Between Parallel Plates

When two parallel conducting plates are connected to a constant potential difference $V$, a nearly uniform electric field forms between the plates (edge effects are ignored for most exam problems). Field strength is constant across all points between the plates.

**Uniform field strength** — For parallel plates separated by distance $d$, field strength is given by $E = \frac{V}{d}$, where $V$ is the potential difference between the plates.

*Notation:* E

**Worked example:** Two parallel plates separated by 5.0 mm are connected to a 12 V battery. Calculate the electric field strength between the plates.

1. Convert separation to SI units: $d = 5.0$ mm $= 5.0 \times 10^{-3}$ m
2. Apply the uniform field formula:
3. $$E = \frac{V}{d} = \frac{12\ \text{V}}{5.0 \times 10^{-3}\ \text{m}} = 2400\ \text{V m}^{-1}$$
4. Since $1$ V m⁻¹ $= 1$ N C⁻¹, the field strength is 2400 N C⁻¹, pointing from the positive plate to the negative plate.

## Motion of Charged Particles in Uniform Fields

A charged particle in a uniform electric field experiences a constant force $F = qE$, so it accelerates at a constant rate $a = \frac{qE}{m}$, where $m$ is the particle mass. When a particle enters the field perpendicular to the field lines, it follows a parabolic projectile path, just like motion in a gravitational field.

**Worked example:** An electron with mass $m_e = 9.1 \times 10^{-31}$ kg and charge $e = 1.6 \times 10^{-19}$ C is in a uniform electric field of 200 N C⁻¹ directed downwards. Calculate the acceleration of the electron.

1. Calculate force on the electron: $F = eE$. The electron is negative, so force acts opposite to the field direction (upwards).
2. Use Newton's second law $a = \frac{F}{m_e}$:
3. $$a = \frac{eE}{m_e} = \frac{(1.6 \times 10^{-19}\ \text{C})(200\ \text{N C}^{-1})}{9.1 \times 10^{-31}\ \text{kg}} \approx 3.5 \times 10^{13}\ \text{m s}^{-2}$$
4. Acceleration is directed upwards, opposite to the field direction.

## Common pitfalls

- **Wrong:** Treating electric field strength as a scalar, adding magnitudes regardless of direction for multiple charges.
  - Why it fails: Electric field is a vector quantity, so direction determines whether fields add or cancel.
  - Correct: Assign directions to each field based on source charge sign, then add as vectors.
- **Wrong:** Using $E = V/d$ for radial point charge fields, or $E = kQ/r^2$ for uniform parallel plate fields.
  - Why it fails: Each formula only applies to its specific type of electric field.
  - Correct: First identify the type of field, then use the matching formula.
- **Wrong:** Taking electric field direction as the direction of force on a negative charge.
  - Why it fails: The definition of electric field uses a positive test charge by convention.
  - Correct: Reverse the direction of force if the charge is negative when finding field direction.
- **Wrong:** Forgetting that the sign of the field value indicates direction, not magnitude.
  - Why it fails: Negative values come from negative source charges, they do not mean the field is smaller.
  - Correct: Report magnitude as an absolute value, and state direction separately if required.

## Cheatsheet

| Concept | Formula | Key Direction Rule |
| --- | --- | --- |
| Field strength (definition) | $E = \frac{F}{q}$ | Same as force on +ve test charge |
| Radial field (point charge) | $E = \frac{kQ}{r^2} = \frac{Q}{4\pi\varepsilon_0 r^2}$ | Away from +Q, towards -Q |
| Uniform field (parallel plates) | $E = \frac{V}{d}$ | From +ve plate to -ve plate |
| Force on charge | $F = qE$ | Same as E for +ve charge, opposite for -ve charge |

## What's next

Mastering electric fields is critical for understanding electric potential, energy changes in electrostatics, and Gauss's law for IB Physics HL. This topic also forms the foundation for analysing combined electric and magnetic fields, which are used to explain phenomena like particle deflection and electromagnetic induction. Strong skills in field calculations will make more advanced topics in the fields theme significantly easier to master.

- [D.3 Motion in electromagnetic fields](https://www.owlsprep.com/study/ib-physics-hl-u4-d-3-motion-in-electromagnetic/)
- [D.4 Magnetic effects of electric currents](https://www.owlsprep.com/study/ib-physics-hl-u4-d-4-magnetic-effects-of/)
- [D.5 Electromagnetic induction (AHL)](https://www.owlsprep.com/study/ib-physics-hl-u4-d-5-electromagnetic-induction/)

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